CODEWITHSUNDEEP
pythonpython-2.7conditional-operator
0x90
0x90

Reputation: 40982

Is there a cleaner and a shorter way to write this conditional in python?

>>> def accept(d1, d2):
    if somefunc(d1,d2) > 32:
        h = 1
    else:
        h = 0
    return h

Does Python have a ternary conditional operator? doesn't give a solution for a case one want to return a value. A lambda based solution is preferable.

Upvotes: 1

Views: 71

Answers (3)

Tim Peters
Tim Peters

Reputation: 70602

Or, perhaps trickier,

return int(somefunc(d1, d2) > 32)

Note that int(True) == 1 and int(False) == 0.

Upvotes: 3

Nacib Neme
Nacib Neme

Reputation: 889

Turn into a lambda (not a explicit function): 

accept = lambda d1,d2: 1 if somefunc(d1, d2) > 32 else 0

Upvotes: -1

BrenBarn
BrenBarn

Reputation: 251368

The "return-value scenario" is no different than any other:

return 1 if somefunc(d1, d2) > 32 else 0

If for some reason you want a lambda:

lambda d1, d2: 1 if somefunc(d1, d2) > 32 else 0

Note that a lambda is no different than a function defined with def that returns the same thing. Lambdas are just regular functions.

Upvotes: 5

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